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Praetorian
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hardworker_indian
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cbrf3
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damit
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e none

a states d-e = +ve perfect sq so d-e could be 9-0 = 9 (sq of 3)
or d-e = 4 (sq of 2)

if d-e = 9 (9,0) then ans .5 if d-e (4,0 or 9-5) = 4 then ans .4 or .5 so not suff

b is in suff becuase sqrt (d) > e * e

if d is 2 sqrt (2) =1 .414

1.4 > e*e then e must be 1

if d is 9 then sqrt (9) = 1.732 > e*e when e = 0 or 1 e cant be 5

combine together still not suff
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hardworker_indian
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cbrf3
Hey hardworker_indian, if d =5 then deimal is 0.451 and if d = 2 decimal is 0.421................Isn't 0.451 rounded to nearest tenth 0.5?? In that case answer should be E......correct me if I am wrong!!
cbrf3


Yes, you are right. Corrected answer:


E.

Number = 0.4 if d<5
Number = 0.5 if d>=5
Our aim is to see if d is lesser than 5.

(1) d – e is equal to a positive perfect square.
combination for positive perfect square for < 10 pairs are:
9 = 9-0
4 = 9-5, 8-4, 7-3, 6-2, 5-1, 4-0
1 = 9-8, 8-7, ...., 2-1
We cannot decide. d may be 2 or 9

(2) sqrt(d) > e^2
d > e^4
e has to be 1, since e=2 will make d>16 not possible.
if e=1, e^4 = 1 and d can be 2 or 9
Insufficient

Together,
e=1 from (2)
d should be 5 or 2 from (1)
In the first case, value = 0.451 ~ 0.5 and
in the second case, value = 0.421 ~ 04
Insufficient.
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Paul
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Yes, I also got E on this one.
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Topgun
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Yea got E...
I think its a challenging question not because its not solvable, but how you solve this under time constraints...
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srijay007
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I got E
d-e can be 1,4,9 however there can be lots of d,e combos with d> or e-square or d>e^4
it is not sufficient for same reason as 1
using 1 and 2 there can the following combos
for e=0 d=1/4/9
for e=1 d=2/5
hence the decimal can be rounded off to either 0.5(e0d9) or 0.4(for other combos



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